YES

We show the termination of the TRS R:

  active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
  mark(f(X1,X2)) -> active(f(mark(X1),X2))
  mark(g(X)) -> active(g(mark(X)))
  f(mark(X1),X2) -> f(X1,X2)
  f(X1,mark(X2)) -> f(X1,X2)
  f(active(X1),X2) -> f(X1,X2)
  f(X1,active(X2)) -> f(X1,X2)
  g(mark(X)) -> g(X)
  g(active(X)) -> g(X)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(f(g(X),Y)) -> mark#(f(X,f(g(X),Y)))
p2: active#(f(g(X),Y)) -> f#(X,f(g(X),Y))
p3: mark#(f(X1,X2)) -> active#(f(mark(X1),X2))
p4: mark#(f(X1,X2)) -> f#(mark(X1),X2)
p5: mark#(f(X1,X2)) -> mark#(X1)
p6: mark#(g(X)) -> active#(g(mark(X)))
p7: mark#(g(X)) -> g#(mark(X))
p8: mark#(g(X)) -> mark#(X)
p9: f#(mark(X1),X2) -> f#(X1,X2)
p10: f#(X1,mark(X2)) -> f#(X1,X2)
p11: f#(active(X1),X2) -> f#(X1,X2)
p12: f#(X1,active(X2)) -> f#(X1,X2)
p13: g#(mark(X)) -> g#(X)
p14: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1, p3, p5, p6, p8}
  {p9, p10, p11, p12}
  {p13, p14}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(f(g(X),Y)) -> mark#(f(X,f(g(X),Y)))
p2: mark#(g(X)) -> mark#(X)
p3: mark#(g(X)) -> active#(g(mark(X)))
p4: mark#(f(X1,X2)) -> mark#(X1)
p5: mark#(f(X1,X2)) -> active#(f(mark(X1),X2))

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    active#_A(x1) = x1 + 4
    f_A(x1,x2) = 10
    g_A(x1) = 5
    mark#_A(x1) = 14
    mark_A(x1) = max{0, x1 - 1}
    active_A(x1) = max{2, x1 - 1}

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(f(g(X),Y)) -> mark#(f(X,f(g(X),Y)))
p2: mark#(g(X)) -> mark#(X)
p3: mark#(f(X1,X2)) -> mark#(X1)
p4: mark#(f(X1,X2)) -> active#(f(mark(X1),X2))

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(f(g(X),Y)) -> mark#(f(X,f(g(X),Y)))
p2: mark#(f(X1,X2)) -> active#(f(mark(X1),X2))
p3: mark#(f(X1,X2)) -> mark#(X1)
p4: mark#(g(X)) -> mark#(X)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    active#_A(x1) = x1 + 7
    f_A(x1,x2) = max{11, x1 + 1, x2 - 10}
    g_A(x1) = x1 + 17
    mark#_A(x1) = max{25, x1 + 11}
    mark_A(x1) = x1 + 4
    active_A(x1) = max{4, x1}

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(f(g(X),Y)) -> mark#(f(X,f(g(X),Y)))
p2: mark#(f(X1,X2)) -> active#(f(mark(X1),X2))
p3: mark#(f(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(f(g(X),Y)) -> mark#(f(X,f(g(X),Y)))
p2: mark#(f(X1,X2)) -> mark#(X1)
p3: mark#(f(X1,X2)) -> active#(f(mark(X1),X2))

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    active#_A(x1) = x1 + 7
    f_A(x1,x2) = max{3, x1, x2 - 5}
    g_A(x1) = max{10, x1 + 5}
    mark#_A(x1) = x1 + 11
    mark_A(x1) = x1 + 4
    active_A(x1) = max{1, x1}

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(f(X1,X2)) -> mark#(X1)
p2: mark#(f(X1,X2)) -> active#(f(mark(X1),X2))

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(f(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = x1
    f_A(x1,x2) = max{x1 + 1, x2 + 1}

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(mark(X1),X2) -> f#(X1,X2)
p2: f#(X1,active(X2)) -> f#(X1,X2)
p3: f#(active(X1),X2) -> f#(X1,X2)
p4: f#(X1,mark(X2)) -> f#(X1,X2)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = x2
    mark_A(x1) = x1 + 1
    active_A(x1) = x1

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(mark(X1),X2) -> f#(X1,X2)
p2: f#(X1,active(X2)) -> f#(X1,X2)
p3: f#(active(X1),X2) -> f#(X1,X2)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(mark(X1),X2) -> f#(X1,X2)
p2: f#(active(X1),X2) -> f#(X1,X2)
p3: f#(X1,active(X2)) -> f#(X1,X2)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = x2 + 1
    mark_A(x1) = max{0, x1 - 1}
    active_A(x1) = x1 + 1

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(mark(X1),X2) -> f#(X1,X2)
p2: f#(active(X1),X2) -> f#(X1,X2)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(mark(X1),X2) -> f#(X1,X2)
p2: f#(active(X1),X2) -> f#(X1,X2)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = x1
    mark_A(x1) = x1
    active_A(x1) = x1 + 1

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(mark(X1),X2) -> f#(X1,X2)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(mark(X1),X2) -> f#(X1,X2)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = x1
    mark_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(mark(X)) -> g#(X)
p2: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    g#_A(x1) = max{0, x1 - 1}
    mark_A(x1) = x1 + 1
    active_A(x1) = x1 + 2

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(mark(X)) -> g#(X)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(mark(X)) -> g#(X)

and R consists of:

r1: active(f(g(X),Y)) -> mark(f(X,f(g(X),Y)))
r2: mark(f(X1,X2)) -> active(f(mark(X1),X2))
r3: mark(g(X)) -> active(g(mark(X)))
r4: f(mark(X1),X2) -> f(X1,X2)
r5: f(X1,mark(X2)) -> f(X1,X2)
r6: f(active(X1),X2) -> f(X1,X2)
r7: f(X1,active(X2)) -> f(X1,X2)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    g#_A(x1) = x1
    mark_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.